# Unitary Operator on a Tensor Product

1. Apr 7, 2010

### barnflakes

Having a little trouble deriving a result in a book.

If I have an operator of the form $$e^{\alpha A \otimes I_n}$$

Where alpha is a complex constant, A a square hermitian matrix and I the identity matrix.

Now if I want to operator that on a tensor product, say for instance $$c_{n,1} |1 \rangle \otimes |n \rangle$$ then how would I do that?

I firstly used the identity $$e^{\alpha A \otimes I_n} = e^{\alpha A \otimes I_n}$$ to obtain:
$$e^{\alpha A \otimes I_n} {} | c_{n,1} |1 \rangle \otimes |n \rangle \rangle = e^{\alpha A} \otimes I_n {} |c_{n,1} {}|1\rangle \otimes |n \rangle \rangle = c_{n,1} e^{\alpha A}{}|1\rangle \otimes |n \rangle$$

but my book gets $$c_{n,1} e^{\alpha}|1 \rangle \otimes |n \rangle$$ with no further explanation. By the way the form of the matrix $$A = \begin{pmatrix} 1 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & -1 \end{pmatrix}$$ if it helps to know it.

Also I am aware you can represent $$e^{\alpha A}$$ in the form of an infinite series but I don't see how that helps here. In fact I tried it and I didn't know where I should cut the series off at, and it gave me coefficients of alpha rather than e^alpha. Oh and the n's are being summed over, not sure if that makes any difference.

Last edited: Apr 7, 2010
2. Apr 7, 2010

### Fredrik

Staff Emeritus
$$e^{A\otimes B}|\alpha\rangle\otimes|\beta\rangle=\sum_{n=0}^\infty \frac{1}{n!}(A\otimes B)^n|\alpha\rangle\otimes|\beta\rangle=\sum_{n=0}^\infty \frac{1}{n!}\underbrace{(A\otimes B)\cdots(A\otimes B)}_{n\text{ factors}}|\alpha\rangle\otimes|\beta\rangle$$

$$=\sum_{n=0}^\infty \frac{1}{n!}\underbrace{(A\otimes B)\cdots(A\otimes B)}_{n-1\text{ factors}}A|\alpha\rangle\otimes B|\beta\rangle=\dots=\sum_{n=0}^\infty \frac{1}{n!}A^n|\alpha\rangle\otimes B^n|\beta\rangle$$

Now what do you get when you set B=I?

3. Apr 8, 2010

### barnflakes

Using your definition $$\sum_{n=0}^\infty \frac{1}{n!}A^n|\alpha\rangle\otimes B^n|\beta\rangle$$ applied to the above expression, I obtain:

$$\sum_{k=0}^\infty \frac{1}{k!}(\alpha A)^k|1\rangle\otimes|n\rangle$$

I cannot quite see how to turn that into $$e^{\alpha}|1 \rangle \otimes |n \rangle$$

What happens to the matrix A? The vector $$|1\rangle$$ is an eigenvector of A with eigenvalue 1.

So can I express the above as:

$$\sum_{k=0}^\infty \frac{1}{k!}(\alpha A)^k|1\rangle \otimes|n\rangle = \sum_{k=0}^\infty \frac{1}{k!}(\alpha)^k A^k|1\rangle \otimes|n \rangle$$ and then work out $$A^k |1\rangle$$

However, I don't know how to work out $$A^k |1\rangle$$ ? I know that $$A |1\rangle = 1 |1\rangle = |1\rangle$$ but how can I use this to calculate the action of A^k?

4. Apr 8, 2010

### barnflakes

Hold on, I have solved it!

$$A^k = A.A...A$$ k times therefore $$A^k |1 \rangle = |1\rangle$$

Thank you Fredrik :)